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  • Python : How to add month to December 2012 and get January 2013?

    - by daydreamer
    >>> start_date = date(1983, 11, 23) >>> start_date.replace(month=start_date.month+1) datetime.date(1983, 12, 23) This works until the month is <=11, as soon as I do >>> start_date = date(1983, 12, 23) >>> start_date.replace(month=start_date.month+1) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: month must be in 1..12 How can I keep adding months which increments the year when new month is added to December?

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  • How to display a page in my browser with python code that is run locally on my computer with "GAE" S

    - by brilliant
    When I run this code on my computer with the help of "Google App Engine SDK", it displays (in my browser) the HTML code of the Google home page: from google.appengine.api import urlfetch url = "http://www.google.com/" result = urlfetch.fetch(url) print result.content How can I make it display the page itself? I mean I want to see that page in my browser the way it would normally be seen by any user of the internet.

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  • (Python/Pyramid) Better ways to have standard list/form editors?

    - by badcat
    I'm working on a number of Pyramid (former Pylons) projects, and often I have the need to display a list of some content (let's say user accounts, log entries or simply some other data). A user should be able to paginate through the list, click on a row and get a form where he/she can edit the contents of that row. Right now I'm always re-inventing the wheel by having Mako templates which use webhelpers for the pagination, Jquery UI for providing a dialog and I craft the editor form and AJAX requests on the client and server side by hand. As you may know, this eats up painfully much time. So what I'm wondering is: Is there a better way of providing lists, editor dialog and server/client communication about this, without having to re-invent the wheel every time? I heard Django takes off a big load of that by providing user accounts and other stuff out of the box; but in my case it's not just about user accounts, it can be any kind of data that is stored on the server-side in a SQL database, which should be able to be edited by a user. Thanks in advance!

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  • How to test that variable is not equal to multiple things? Python

    - by M830078h
    This is the piece of code I have: choice = "" while choice != "1" and choice != "2" and choice != "3": choice = raw_input("pick 1, 2 or 3") if choice == "1": print "1 it is!" elif choice == "2": print "2 it is!" elif choice == "3": print "3 it is!" else: print "You should choose 1, 2 or 3" While it works, I feel that it's really clumsy, specifically the while clause. What if I have more acceptable choices? Is there a better way to make the clause?

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  • Are python list comprehensions always a good programming practice?

    - by dln385
    To make the question clear, I'll use a specific example. I have a list of college courses, and each course has a few fields (all of which are strings). The user gives me a string of search terms, and I return a list of courses that match all of the search terms. This can be done in a single list comprehension or a few nested for loops. Here's the implementation. First, the Course class: class Course: def __init__(self, date, title, instructor, ID, description, instructorDescription, *args): self.date = date self.title = title self.instructor = instructor self.ID = ID self.description = description self.instructorDescription = instructorDescription self.misc = args Every field is a string, except misc, which is a list of strings. Here's the search as a single list comprehension. courses is the list of courses, and query is the string of search terms, for example "history project". def searchCourses(courses, query): terms = query.lower().strip().split() return tuple(course for course in courses if all( term in course.date.lower() or term in course.title.lower() or term in course.instructor.lower() or term in course.ID.lower() or term in course.description.lower() or term in course.instructorDescription.lower() or any(term in item.lower() for item in course.misc) for term in terms)) You'll notice that a complex list comprehension is difficult to read. I implemented the same logic as nested for loops, and created this alternative: def searchCourses2(courses, query): terms = query.lower().strip().split() results = [] for course in courses: for term in terms: if (term in course.date.lower() or term in course.title.lower() or term in course.instructor.lower() or term in course.ID.lower() or term in course.description.lower() or term in course.instructorDescription.lower()): break for item in course.misc: if term in item.lower(): break else: continue break else: continue results.append(course) return tuple(results) That logic can be hard to follow too. I have verified that both methods return the correct results. Both methods are nearly equivalent in speed, except in some cases. I ran some tests with timeit, and found that the former is three times faster when the user searches for multiple uncommon terms, while the latter is three times faster when the user searches for multiple common terms. Still, this is not a big enough difference to make me worry. So my question is this: which is better? Are list comprehensions always the way to go, or should complicated statements be handled with nested for loops? Or is there a better solution altogether?

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  • Best way in Python to determine all possible intersections in a matrix?

    - by ssweens
    So if I have a matrix (list of lists) of unique words as my column headings, document ids as my row headings, and a 0 or 1 as the values if the word exists in that particular document. What I'd like to know is how to determine all the possible combinations of words and documents where more than one word is in common with more than one document. So something like: [[Docid_3, Docid_5], ['word1', 'word17', 'word23']], [[Docid_3, Docid_9, Docid_334], ['word2', 'word7', 'word23', 'word68', 'word982']], and so on for each possible combination. Would love a solution that provides the complete set of combinations and one that yields only the combinations that are not a subset of another, so from the example, not [[Docid_3, Docid_5], ['word1', 'word17']] since it's a complete subset of the first example. I feel like there is an elegant solution that just isn't coming to mind and the beer isn't helping. Thanks.

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  • The return value should be a list but doesn't return as expected?! - Python newbie

    - by user1432941
    Hi this must be a very simple solution that has eluded me this last hour. I've tried to build this test function where the return value of the test_cases list should match the values in the test_case_answers list but for some reason, test case 1 and test case 2 fail. When i print the return values for the test cases they return the correct answers, but for some reason test case 1 and test case 2 return False. Thanks for your help! import math test_cases = [1, 9, -3] test_case_answers = [1, 3, 0] def custom_sqrt(num): for i in range(len(test_cases)): if test_cases[i] >= 0: return math.sqrt(test_cases[i]) else: return 0 for i in range(len(test_cases)): if custom_sqrt(test_cases[i]) != test_case_answers[i]: print "Test Case #", i, "failed!" custom_sqrt(test_cases)

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  • Python - Find where in string regex match was found?

    - by nb
    I'm currently using regular expressions to search through RSS feeds to find if certain words and phrases are mentioned, and would then like to extract the text on either side of the match as well. For example: String = "This is an example sentence, it is for demonstration only" re.search("is", String) I'd like to know where the is was found so that I can extract and output something like this: 1 match found: "This is an example sentence" I know that it would be easy to do with splits, but I'd need to know what the index of first character of the match was in the string, which I don't know how to find

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  • Gtk_entry mouse click signal

    - by h4rp0
    Hi, I want to do a g_signal_connect to capture a mouse click in a gtk_entry widget. Something like this: entry = gtk_entry_new ( ); gtk_box_pack_end ( GTK_BOX ( hBox ), entry, TRUE, TRUE, 1 ); gtk_widget_show ( entry ); // This is the one I'm not sure about g_signal_connect ( GTK_OBJECT ( entry ), "????????????", GTK_SIGNAL_FUNC ( EntryClicked ), entry ); I just can't seem to find it in the gtk documentation. I've tried using the "focus-in-event", but it is not working as I spect. Thanks for the assistance.

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  • program received signal SIGABRT (xcode)

    - by manish1990
    #import <UIKit/UIKit.h> @interface tableview : UIViewController<UITableViewDataSource> { NSArray *listOfItems; } @property(nonatomic,retain) NSArray *listOfItems; @end #import "tableview.h" @implementation tableview @synthesize listOfItems; - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *CellIdentifier = @"Cell"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier ]autorelease]; } //NSString *cellValue = [listOfItems objectAtIndex:indexPath.row]; cell.textLabel.text = [listOfItems objectAtIndex:indexPath.row]; return cell; } - (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section { return 3; } - (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil { self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil]; if (self) { // Custom initialization } return self; } - (void)didReceiveMemoryWarning { // Releases the view if it doesn't have a superview. [super didReceiveMemoryWarning]; // Release any cached data, images, etc that aren't in use. } #pragma mark - View lifecycle - (void)viewDidLoad { listOfItems = [[NSArray alloc] initWithObjects:@"first",@"second",@"third", nil]; //listOfItems = [[NSMutableArray alloc]init]; // [listOfItems addObject:@"first"]; //[listOfItems addObject:@"second"]; [super viewDidLoad]; // Do any additional setup after loading the view from its nib. } -(void)dealloc { [listOfItems release]; [super dealloc]; } @end GNU gdb 6.3.50-20050815 (Apple version gdb-1708) (Mon Aug 15 16:03:10 UTC 2011) Copyright 2004 Free Software Foundation, Inc. GDB is free software, covered by the GNU General Public License, and you are welcome to change it and/or distribute copies of it under certain conditions. Type "show copying" to see the conditions. There is absolutely no warranty for GDB. Type "show warranty" for details. This GDB was configured as "x86_64-apple-darwin".sharedlibrary apply-load-rules all Attaching to process 438. 2012-04-27 13:33:23.276 tableview test[438:207] -[UIView tableView:numberOfRowsInSection:]: unrecognized selector sent to instance 0x6855500 2012-04-27 13:33:23.362 tableview test[438:207] * Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[UIView tableView:numberOfRowsInSection:]: unrecognized selector sent to instance 0x6855500' * First throw call stack: (0x13bb052 0x154cd0a 0x13bcced 0x1321f00 0x1321ce2 0x1ecf2b 0x1ef722 0x9f7c7 0x9f2c1 0xa228c 0xa6783 0x51322 0x13bce72 0x1d6592d 0x1d6f827 0x1cf5fa7 0x1cf7ea6 0x1d8330c 0x23530 0x138f9ce 0x1326670 0x12f24f6 0x12f1db4 0x12f1ccb 0x12a4879 0x12a493e 0x12a9b 0x2282 0x21f5) terminate called throwing an exceptionCurrent language: auto; currently objective-c (gdb)

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  • Python raises a KeyError (for an out of dictionary key) even though the key IS in the dictionary

    - by ignorantslut
    I'm getting a KeyError for an out of dictionary key, even though I know the key IS in fact in the dictionary. Any ideas as to what might be causing this? print G.keys() returns the following: ['24', '25', '20', '21', '22', '23', '1', '3', '2', '5', '4', '7', '6', '9', '8', '11', '10', '13', '12', '15', '14', '17', '16', '19', '18'] but when I try to access a value in the dictionary on the next line of code... for w in G[v]: #note that in this example, v = 17 I get the following error message: KeyError: 17 Any help, tips, or advice are all appreciated. Thanks.

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  • Python - Is a dictionary slow to find frequency of each character?

    - by psihodelia
    I am trying to find a frequency of each symbol in any given text using an algorithm of O(n) complexity. My algorithm looks like: s = len(text) P = 1.0/s freqs = {} for char in text: try: freqs[char]+=P except: freqs[char]=P but I doubt that this dictionary-method is fast enough, because it depends on the underlying implementation of the dictionary methods. Is this the fastest method?

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  • How to select random image of specific size using Django / Python?

    - by Jonathan
    I've been using this little snippet to select random images. However I would like to change it to select only images of a certain size. I'm running into trouble checking against image size. If I use get_image_dimensions() I need to use a conditional statement, which then requires that I allow exceptions. So, I guess I need some pointers on just limiting by image dimensions. Thanks. import os import random import posixpath from django import template from django.conf import settings register = template.Library() def is_image_file(filename): """Does `filename` appear to be an image file?""" img_types = [".jpg", ".jpeg", ".png", ".gif"] ext = os.path.splitext(filename)[1] return ext in img_types @register.simple_tag def random_image(path): """ Select a random image file from the provided directory and return its href. `path` should be relative to MEDIA_ROOT. Usage: <img src='{% random_image "images/whatever/" %}'> """ fullpath = os.path.join(settings.MEDIA_ROOT, path) filenames = [f for f in os.listdir(fullpath) if is_image_file(f)] pick = random.choice(filenames) return posixpath.join(settings.MEDIA_URL, path, pick)

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